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 Home Brew Forums > On draft serving pressures...of sorts
08-15-2010, 09:02 PM   #1
jgourd
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 On draft serving pressures...of sorts

Hopefully this is the proper forum to ask this. It's more of a scientific question, but it relates to "brewing science" in that I can apply it to dispensing my beer (i.e. draft system).

Suppose I have a 48" length of hose with an inside diameter of 3/16". The hose's supply end is 21.5" below its dispense end. Also suppose that the average flow resistance of such a hose is 2.25 lbs/ft. At the supply end, a pressure of 9.02 PSI is pushing liquid through the hose (assume the liquid is beer with a specific gravity of 1.010 g/cm^3).

1) What is the flow rate of the liquid at the dispense end?
2) What is the pressure (in PSI) at the dispense end?
3) Perhaps the more important question is, what pressure (in PSI) must be present at the dispense end in order to have a desired flow rate of 0.78 gal/min (with all of the specifics of the hose as described above)?

I would prefer formulas as opposed to just, "here's the answer." That way I can learn. Thanks!

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08-16-2010, 04:02 PM   #2
ajdelange
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Quote:
 Originally Posted by jgourd Suppose I have a 48" length of hose with an inside diameter of 3/16". The hose's supply end is 21.5" below its dispense end. Also suppose that the average flow resistance of such a hose is 2.25 lbs/ft. At the supply end, a pressure of 9.02 PSI is pushing liquid through the hose (assume the liquid is beer with a specific gravity of 1.010 g/cm^3). 1) What is the flow rate of the liquid at the dispense end?
This is a practical problem rather than a theoretical one. As you probably know the rate of flow of viscous fluid through a tube depends on the viscosity of the fluid, the diameter and length of the tube and the friction of the fluid with the tube walls. The details are in many physics and fluid mechanics textbooks. The practical approach is to take measurements on actual beverage tubing and it is from these that the various dispense tables are derived. The tables all assume a flow of 2 Oz per second at the faucet (i.e you can fill a 12 Oz glass in 6. Second.)

Quote:
 Originally Posted by jgourd 2) What is the pressure (in PSI) at the dispense end?
Clearly, the pressure at the dispense end is 1 atmosphere (14.7 psia).

Quote:
 Originally Posted by jgourd 3) Perhaps the more important question is, what pressure (in PSI) must be present at the dispense end in order to have a desired flow rate of 0.78 gal/min (with all of the specifics of the hose as described above)?
You'll need enough pressure to overcome the flow resistance (found in tables or from online calculators). Example:http://www.kegworks.com/blog/2007/05...t-beer-system/. Just multiply the total length of the line, including the vertical run, times the resistance figure. e.g. 2 ft 1/4" ID stainless steel: (2)*(1.2) = 2.4 psi; 3 ft 3/16 ID vinyl: (3)*(2.2) = 6.6 psi.

You'll also need the gas to do the work of lifting the beer. The pressure difference at 2 points h apart in a fluid column is simply density*gravitational_constant*h. It works out quite simply here as an atmosphere is 30 feet of water and 14.7 psia. So it takes about 0.5 psi per foot to nullify the hydrostatic pressure. So the overall formula is

P = sum[i=1,N](resistance_of_hose_i*length_of_hose_i) + 0.5*h

for a system with N bits of hose of different lengths and resistance coefficients.

Note that 0.5*h is for water. For another fluid it would be 0.5*h*SG but as beer has SG so close to 1 and there are so many other approximations (viscosity, for example) it's not worth worrying about. Even with the most accurate formula you will find you have to "tune" the system by tweaking gas pressure (bearing in mind that eventually this will translate into more or less gassy beer), adding "chokers", changing hose diameter or type etc.

I would prefer formulas as opposed to just, "here's the answer." That way I can learn. Thanks![/QUOTE]
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08-16-2010, 09:45 PM   #3
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So if I understand this correctly then my 48" length of 3/16" vinyl tubing (middle of keg is 21.5" below the tap) results in a PSI drop of:

(48"/12 * 2.2 PSI/ft) + (21.5"/12 * 0.5 PSI/ft) ~= 9.7 PSI

And if I set my CO2 regulator to 9.7 PSI, nothing should come out of the tap right? There would be equilibrium. Or would it still come out, but at a very slow rate?

But anyways suppose I want to carbonate my 40F beer at 2.2 volumes. That requires 9 PSI at the regulator (according to the tables). So most people would say to put a tube of an appropriate length to leave you with 1 PSI to serve (i.e. the tube length and keg-to-tap height difference would drop 8 PSI). Problem is, this assumes a constant flow rate out of the tap (something like 2 oz/sec or ~1 gal/min). From what I understand, a beer with more CO2 requires a longer pour. So the flow rate of the beer should change as the CO2 volumes (and therefore PSI) changes. I'm interested in finding some sort of correlation between PSI and flow rate. So if I want a 0.78 gal/min flow rate out of the tap, how many PSI's does that equate to given the tubing specs? Said differently, how much of a PSI drop would I need in order to have enough pressure left over to equate to an 0.78 gal/min flow rate? For 3/16" tubing, maybe 1 PSI equals 1 gal/min flow rate; I don't know.

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08-17-2010, 04:29 AM   #4
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Quote:
 Originally Posted by jgourd So if I understand this correctly then my 48" length of 3/16" vinyl tubing (middle of keg is 21.5" below the tap) results in a PSI drop of: (48"/12 * 2.2 PSI/ft) + (21.5"/12 * 0.5 PSI/ft) ~= 9.7 PSI
Yes.

Quote:
 Originally Posted by jgourd And if I set my CO2 regulator to 9.7 PSI, nothing should come out of the tap right? There would be equilibrium. Or would it still come out, but at a very slow rate?
It would come out at 2 Oz per second. The resistance per foot is measured at that flow rate. And yes, there would be an equilibrium: a steady state flow at 2 Oz/sec.

Quote:
 Originally Posted by jgourd But anyways suppose I want to carbonate my 40F beer at 2.2 volumes. That requires 9 PSI at the regulator (according to the tables). So most people would say to put a tube of an appropriate length to leave you with 1 PSI to serve (i.e. the tube length and keg-to-tap height difference would drop 8 PSI).
I think most people would say to put on a tube length that gave you 9 PSI drop.

Quote:
 Originally Posted by jgourd Problem is, this assumes a constant flow rate out of the tap (something like 2 oz/sec or ~1 gal/min).
Yes.

Quote:
 Originally Posted by jgourd From what I understand, a beer with more CO2 requires a longer pour. So the flow rate of the beer should change as the CO2 volumes (and therefore PSI) changes.
I've never heard that but if you want to slow the pour somewhat you can simply increase the line length a bit (add a foot of choker).

Quote:
 Originally Posted by jgourd I'm interested in finding some sort of correlation between PSI and flow rate.
For laminar flow the rate is proportional to the pressure. For turbulent flow, to the square root of pressure. The flow is probably mostly laminar but not completely so you're probably close to, but not quite linear.

Quote:
 Originally Posted by jgourd So if I want a 0.78 gal/min flow rate out of the tap, how many PSI's does that equate to given the tubing specs? Said differently, how much of a PSI drop would I need in order to have enough pressure left over to equate to an 0.78 gal/min flow rate?
The pressure drop is P =8*u* L*Q/pi*r^4 where u is the viscosity, L is the length Q the flow and r the radius. If you keep the same tubing you can slow the flow to 100 Oz/min (0.78 GPM) simply by decreasing the pressure by the ratio of the flows i.e. if 9 psi gives you 120 Oz/min then 9*100/120 would give you 100. But the lower pressure would result in lower carbonation but it would take weeks for re-equilibration so turning the gas down a bit for serving and then restoring it at the end of the session is certainly a feasible approach. The "right" answer, however, is to leave the pressure at 9 and adjust L up by the same factor you want to decrease Q i.e. you should change the tube length to 120/100 = 1.2 times what you would have for 120 Oz/min. This maintains the pressure drop across the line at 9 psi at the lower flow. All this is approximate, of course, so experimentation is usually required to get things just right.
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08-17-2010, 01:58 PM   #5
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Quote:
 Originally Posted by ajdelange It would come out at 2 Oz per second. The resistance per foot is measured at that flow rate. And yes, there would be an equilibrium: a steady state flow at 2 Oz/sec.
I understand now. The line resistance tables are designed with the 1 gpm flow rate in mind.

Quote:
 Originally Posted by ajdelange I think most people would say to put on a tube length that gave you 9 PSI drop.
Got it.

Quote:
 Originally Posted by ajdelange I've never heard that but if you want to slow the pour somewhat you can simply increase the line length a bit (add a foot of choker).
See this document for an explanation of this idea.

Quote:
 Originally Posted by ajdelange For laminar flow the rate is proportional to the pressure. For turbulent flow, to the square root of pressure. The flow is probably mostly laminar but not completely so you're probably close to, but not quite linear. The pressure drop is P =8*u* L*Q/pi*r^4 where u is the viscosity, L is the length Q the flow and r the radius. If you keep the same tubing you can slow the flow to 100 Oz/min (0.78 GPM) simply by decreasing the pressure by the ratio of the flows i.e. if 9 psi gives you 120 Oz/min then 9*100/120 would give you 100. But the lower pressure would result in lower carbonation but it would take weeks for re-equilibration so turning the gas down a bit for serving and then restoring it at the end of the session is certainly a feasible approach. The "right" answer, however, is to leave the pressure at 9 and adjust L up by the same factor you want to decrease Q i.e. you should change the tube length to 120/100 = 1.2 times what you would have for 120 Oz/min. This maintains the pressure drop across the line at 9 psi at the lower flow. All this is approximate, of course, so experimentation is usually required to get things just right.
I agree that the solution is to alter the length of the tubing in order to increase/decrease the PSI drop. Thanks for the clarification and "lesson!" I'll think about it and see how it might all apply later today.
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08-17-2010, 05:04 PM   #6
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The document was interesting but didn't give much by way of explanation as to why one would want different flows for different beers much beyond anyone who tells you that 2 Oz/sec is the right number is trying to sell you something (and the writer isn't!). I found his tabulated data on resistance interesting. The relationship between flow and resistance shows a 1.8 power dependence which says that the flow is not laminar. That surprises me a little. Doesn't seem as if a couple of ounces a sec. is that much but fluids isn't my bag by any means. IOW I'm not questioning the data - just commenting on it. The other interesting thing from the table is that pressure drop as a function of diameter shows a 5th power dependence as opposed to 4th in the Hagen-Poiseuille equation (posted last night) for laminar flow.

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